What is the maximal domain of these functions?












0












$begingroup$


Need help figuring out, how to find the maximal domains of functions



$f(x) = frac{x^3-x}{2x^2+1}$



$f(x) = frac{x^2-5x+6}{x-2}$






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
    $endgroup$
    – Piwi
    Oct 11 '15 at 17:28












  • $begingroup$
    please clarify the domain of function argument
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:18
















0












$begingroup$


Need help figuring out, how to find the maximal domains of functions



$f(x) = frac{x^3-x}{2x^2+1}$



$f(x) = frac{x^2-5x+6}{x-2}$






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
    $endgroup$
    – Piwi
    Oct 11 '15 at 17:28












  • $begingroup$
    please clarify the domain of function argument
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:18














0












0








0





$begingroup$


Need help figuring out, how to find the maximal domains of functions



$f(x) = frac{x^3-x}{2x^2+1}$



$f(x) = frac{x^2-5x+6}{x-2}$






functions







share|cite|improve this question











$endgroup$




Need help figuring out, how to find the maximal domains of functions



$f(x) = frac{x^3-x}{2x^2+1}$



$f(x) = frac{x^2-5x+6}{x-2}$






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 4 '18 at 13:11









pointguard0

1,4581021




1,4581021










asked Oct 11 '15 at 17:05









Chris SmithChris Smith

611




611












  • $begingroup$
    I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
    $endgroup$
    – Piwi
    Oct 11 '15 at 17:28












  • $begingroup$
    please clarify the domain of function argument
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:18


















  • $begingroup$
    I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
    $endgroup$
    – Piwi
    Oct 11 '15 at 17:28












  • $begingroup$
    please clarify the domain of function argument
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:18
















$begingroup$
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
$endgroup$
– Piwi
Oct 11 '15 at 17:28






$begingroup$
I guess $f$ should be dependant on $y$ and $z$, not on $x$? Considering finding the domain: The domain of $f$ is the set of arguments for which $f$ is well-defined. So you need to ask yourself "when is $frac{a}{b}$ not well-defined?".
$endgroup$
– Piwi
Oct 11 '15 at 17:28














$begingroup$
please clarify the domain of function argument
$endgroup$
– pointguard0
Aug 4 '18 at 13:18




$begingroup$
please clarify the domain of function argument
$endgroup$
– pointguard0
Aug 4 '18 at 13:18










2 Answers
2






active

oldest

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0












$begingroup$

I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.




  1. This is a rational expression. The only restriction is that the denominator is not $0$. Since
    $$2y^2+1=0$$
    has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.


  2. Same thinking leads to
    $$z-2ne 0 Leftrightarrow zne 2.$$
    The domain is $mathbb{R}backslash{2}$.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    function argument could be complex.
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:04



















0












$begingroup$

The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.




    1. This is a rational expression. The only restriction is that the denominator is not $0$. Since
      $$2y^2+1=0$$
      has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.


    2. Same thinking leads to
      $$z-2ne 0 Leftrightarrow zne 2.$$
      The domain is $mathbb{R}backslash{2}$.







    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      function argument could be complex.
      $endgroup$
      – pointguard0
      Aug 4 '18 at 13:04
















    0












    $begingroup$

    I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.




    1. This is a rational expression. The only restriction is that the denominator is not $0$. Since
      $$2y^2+1=0$$
      has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.


    2. Same thinking leads to
      $$z-2ne 0 Leftrightarrow zne 2.$$
      The domain is $mathbb{R}backslash{2}$.







    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      function argument could be complex.
      $endgroup$
      – pointguard0
      Aug 4 '18 at 13:04














    0












    0








    0





    $begingroup$

    I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.




    1. This is a rational expression. The only restriction is that the denominator is not $0$. Since
      $$2y^2+1=0$$
      has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.


    2. Same thinking leads to
      $$z-2ne 0 Leftrightarrow zne 2.$$
      The domain is $mathbb{R}backslash{2}$.







    share|cite|improve this answer









    $endgroup$



    I will suppose $f(x)$ should be $f(y)$ and $f(z)$ and that we are working in $mathbb{R}$.




    1. This is a rational expression. The only restriction is that the denominator is not $0$. Since
      $$2y^2+1=0$$
      has no solution then the denominator does not impose any restriction on the domain. Hence $dom(f)=mathbb{R}$.


    2. Same thinking leads to
      $$z-2ne 0 Leftrightarrow zne 2.$$
      The domain is $mathbb{R}backslash{2}$.








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 11 '15 at 22:22









    Jean-François GagnonJean-François Gagnon

    897315




    897315












    • $begingroup$
      function argument could be complex.
      $endgroup$
      – pointguard0
      Aug 4 '18 at 13:04


















    • $begingroup$
      function argument could be complex.
      $endgroup$
      – pointguard0
      Aug 4 '18 at 13:04
















    $begingroup$
    function argument could be complex.
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:04




    $begingroup$
    function argument could be complex.
    $endgroup$
    – pointguard0
    Aug 4 '18 at 13:04











    0












    $begingroup$

    The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.






        share|cite|improve this answer









        $endgroup$



        The second function does not contain $z=2$ in the domain, but it can be extended continuously at this point, thus defined continuous, so the maximal domain would be all real numbers.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 11 '15 at 23:47









        user279325user279325

        17416




        17416






























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